Difference between revisions of "2005 AMC 10B Problems/Problem 23"

Revision as of 20:21, 26 December 2020

Problem

In trapezoid $ABCD$ we have $\overline{AB}$ parallel to $\overline{DC}$ , $E$ as the midpoint of $\overline{BC}$ , and $F$ as the midpoint of $\overline{DA}$ . The area of $ABEF$ is twice the area of $FECD$ . What is $AB/DC$ ?

$\mathrm{(A)} 2 \qquad \mathrm{(B)} 3 \qquad \mathrm{(C)} 5 \qquad \mathrm{(D)} 6 \qquad \mathrm{(E)} 8$

Solution 1

Since the height of both trapezoids are equal, and the area of $ABEF$ is twice the area of $FECD$ ,

$\frac{AB+EF}{2}=2\left(\frac{DC+EF}{2}\right)$ .

$\frac{AB+EF}{2}=DC+EF$ , so

$AB+EF=2DC+2EF$ .

$EF$ is exactly halfway between $AB$ and $DC$ , so $EF=\frac{AB+DC}{2}$ .

$AB+\frac{AB+DC}{2}=2DC+AB+DC$ , so

$\frac{3}{2}AB+\frac{1}{2}DC=3DC+AB$ , and

$\frac{1}{2}AB=\frac{5}{2}DC$ .

$\frac{AB}{DC} = \boxed{5}$ .

Solution 2

Mark $DC=z$ , $AB=x$ , and $FE=y.$ Note that the heights of trapezoids $ABEF$ & $FECD$ are the same. Mark the height to be $h$ .

Then, we have that $\dfrac{x+y}{2}\cdot h=2(\dfrac{y+z}{2} \cdot h)$ .

From this, we get that $x=2z+y$ .

We also get that $\dfrac{x+z}{2} \cdot h= 3(\dfrac{y+z}{2} \cdot h)$ .

Simplifying, we get that $2x=z+3y$

Notice that we want $\dfrac{AB}{DC}=\frac{x}{z}$ .

Dividing the first equation by $z$ , we get that $\dfrac{x}{z}=2+\dfrac{y}{z}\implies 3(\dfrac{x}{z})=6+3(\dfrac{y}{z})$ .

Dividing the second equation by $z$ , we get that $2(\dfrac{x}{z})=1+3(\dfrac{y}{z})$ .

Now, when we subtract the top equation from the bottom, we get that $\dfrac{x}{z}=5$

Hence, the answer is $\boxed{5}$

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 28: / Line 28: @@
 Note that the heights of trapezoids <math>ABEF</math> & <math>FECD</math> are the same. Mark the height to be <math>h</math>.
-Then, we have that <math>\frac{x+y}{2}\cdot h=2(\frac{y+z}{2} \cdot h)</math>.
+Then, we have that <math>\dfrac{x+y}{2}\cdot h=2(\dfrac{y+z}{2} \cdot h)</math>.
 From this, we get that <math>x=2z+y</math>.
-We also get that <math>\frac{x+z}{2} \cdot h= 3(\frac{y+z}{2} \cdot h)</math>.
+We also get that <math>\dfrac{x+z}{2} \cdot h= 3(\dfrac{y+z}{2} \cdot h)</math>.
 Simplifying, we get that <math>2x=z+3y</math>
-Notice that we want <math>\frac{AB}{DC}=\frac{x}{z}</math>.
+Notice that we want <math>\dfrac{AB}{DC}=\frac{x}{z}</math>.
-Dividing the first equation by <math>z</math>, we get that <math>\frac{x}{z}=2+\frac{y}{z}\implies 3(\frac{x}{z})=6+3(\frac{y}{z})</math>.
+Dividing the first equation by <math>z</math>, we get that <math>\dfrac{x}{z}=2+\dfrac{y}{z}\implies 3(\dfrac{x}{z})=6+3(\dfrac{y}{z})</math>.
-Dividing the second equation by <math>z</math>, we get that <math>2(\frac{x}{z})=1+3(\frac{y}{z})</math>.
+Dividing the second equation by <math>z</math>, we get that <math>2(\dfrac{x}{z})=1+3(\dfrac{y}{z})</math>.
-Now, when we subtract the top equation from the bottom, we get that <math>\frac{x}{z}=5</math>
+Now, when we subtract the top equation from the bottom, we get that <math>\dfrac{x}{z}=5</math>
 Hence, the answer is <math>\boxed{5}</math>

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Difference between revisions of "2005 AMC 10B Problems/Problem 23"

Revision as of 20:21, 26 December 2020

Contents

Problem

Solution 1

Solution 2

See Also